31 hungry lions! What will happen if...?

Question

There are 31 hungry lions on an island. On the island, there is one piece of meat - however, it is laced with sleeping powder. Thus, if a lion eats this piece of meat, he will fall asleep, and then is liable to be eaten. If a another lion eats this lion who fell asleep, he will also fall asleep (and the cycle continues).

The lions are all trained logicians, and will only eat another lion if they are asleep. They would all rather starve to death than be eaten.

What happens?

Assume there is no other food. There is no sharing allowed (only one lion can eat a piece of meat or another lion. Assume they they only fall asleep when they eat.

I'll answer any questions/clarifications through additional details.

Answer 1

The meat is eaten by a lion, and all the other lions starve.

EDIT:

Explanation: A lion will eat the meat because he knows he won't be eaten when he falls asleep.  To see how this works, we can treat it as a reduction problem:

If there is only 1 lion, he will obviously eat the meat.

If there are 2 lions, they will obviously starve.

Now, if there are 3 lions, and one eats the meat and falls asleep - it essentially becomes the 2 lion scenario.  Thus, the one that eats the meat knows the other two will not eat him and instead starve.

With 4 lions, if one lion eats the meat, then it becomes the 3 lion scenario, and he will be eaten.  So no lions will eat, and all will starve.

Basically, if the number of lions is odd, then a lion will eat the meat because he knows the others will not eat him.

If there are an even number of lions, they will all starve.

^_^

Answer 2

I was there, and I was the quickest to realize that, oddly enough, it's safe to eat the meat, since anyone else foolish enough to eat me would be eaten in turn.  The others starved to death, I barely survived.

You can solve this by thinking successively what would happen if there were just 1, 2, 3... lions to start with until you see the pattern.

Answer 3

[EDIT: Ok, I've rethought my answer and revamped it to include formal mathematical induction.]

The key things to remember are this:

1) In any scenario of n lions, one of two things will happen: either the meat gets eaten, or it doesn't get eaten.

2) If the meat never gets eaten, then there will be no sleeping lions.  Since lions can only eat the meat or a sleeping lion, this means that if the meat never gets eaten, then all of the lions will starve.

3) If the meat DOES get eaten, then it's only eaten by one lion, who falls asleep.  If there are initially n lions and one eats the meat, then the problem now becomes exactly like the n-1 where the sleeping lion replaces the meat.

4) For a pride of n lions, all of them are smart enough to know that whoever eats the meat will fall asleep and thus become the new "meat" of the n-1 scenario.  If you have n lions, and they know that the meat gets eaten in the n-1 scenario, then they WON'T touch the meat in their own n lion scenario.  Therefore, WITH AN ISLAND OF N LIONS, THE MEAT GETS EATEN ONLY IF IT DOESN'T GET EATEN FOR N-1 LIONS.  Likewise, IF THE MEAT GETS EATEN ON AN ISLAND WITH N-1 LIONS, THEN AN ISLAND OF N LIONS WILL ALL STARVE.

If n=1, then the lion will eat the meat, knowing that there are no other lions to eat him.

If n=2, then both lions will think "If I eat that meat, then I'll fall asleep, and the other lion will go ahead and eat me, knowing that there's no consequences."  So both lions would choose to abstain and both starve.

Now with n=3, a lion could go eat the meat and know that he won't get eaten, because after eating the meat this becomes the n=2 problem, where the meat isn't touched.

If a pride of n=k lions know that the meat of the n=k-1 problem gets eaten, then they won't eat the meat in the n=k scenario.  If they know the meat doesn't get eaten for n=k-1, then one of them will go ahead and eat the meat, knowing that he can safely fall asleep and not become dinner.  This means the scenarios always alternate between the meat being eaten (and the rest starving) or no meat being eaten (and all starving).

We showed that for n=1, the meat gets eaten.  So in general, if n is even, then all lions willingly starve.  If n is odd, then one lion eats the meat and the rest starve.  Thus for n=31, one lion will eat the meat and the rest will starve.

Answer 4

lion would be hungry

Answer 5

The last lion will wake up after a while, and will be pi55ed off that all his 'friends' are gone, and now he is lonely.

Soon he will be hungry again. If some food doesn't come to visit (like to rescue him) he will die.

Sad lion.

Answer 6

I like Geezah's answer. Well thought out. Of course, it must be assumed that the meat is only enough to fill one lion.

What are the chances of lions falling asleep from exhaustion and weakness while starving?

Post-apocalyptic scenario:

If all the lions are trained logicians, they will starve until the weakest lion falls asleep due to exhaustion. And, after that it will be a feeding frenzy. With the same pattern repeated until there is only one lion left.

But, what about the weakest lion?

He probably thinks to himself that he might as well eat the tainted meat. Since, he will be the first to be eaten anyway. Why not ease the pain by being in a chemical induced sleep.

Answer 7

if they'd rather starve to death than be eaten, then 30 of them would die from starvation and the last surviving one would eat the meat then fall asleep and wake up and continue to eat the rest of the dead lions until that was gone, then the last of the lions would eventually die...

morbid

Answer 8

As "trained logicians" do the lions know that the meat is laced with sleeping powder? That makes a big difference.